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MakcuM [25]
3 years ago
6

21) -5z – 4 + 8 - x; use x = 7, and z = 2

Mathematics
1 answer:
Ivanshal [37]3 years ago
8 0

Answer:

-13

Step-by-step explanation:

-5z-4+8-x

-5(2)-4+8-7

-10-4+8-7

-14+8-7

-21+8

-13

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Carry out three steps of the Bisection Method for f(x)=3x−x4 as follows: (a) Show that f(x) has a zero in [1,2]. (b) Determine w
ELEN [110]

Answer:

a) There's a zero between [1,2]

b) There's a zero between [1.5,2]

c) There's a zero between  [1.5,1.75].

Step-by-step explanation:

We have f(x)=3^x-x^4

A)We need to show that f(x) has a zero in the interval [1, 2]. We have to see if the function f is continuous with f(1) and f(2).

f(x)=3^x-x^4\\\\f(1)=3^1-(1)^4=3-1=2\\\\f(2)=3^2-(2)^4=9-16=(-7)

We can see that f(1) and f(2) have opposite signs. And f(1)>f(2) and the function is continuous, this means that exists a real number c between the interval [1,2] where f(c)=0.

B)We have to repeat the same steps of A)

For the subinterval [1,1.5]:

f(x)=3^x-x^4\\\\f(1)=3^1-(1)^4=3-1=2\\\\f(1.5)=3^1^.^5-(1.5)^4=5.19-5.06=0.13

f(1) and f(1.5) have the same signs, this means there's no zero in the subinterval [1,1.5].

For the subinterval [1.5,2]:

f(x)=3^x-x^4\\\\f(1.5)=3^1^.^5-(1.5)^4=5.19-5.06=0.13\\\\f(2)=3^2-(2)^4=9-16=(-7)

f(1.5) and f(2) have opposite signs, this means there's a zero between the subinterval [1.5,2].

C)We have to repeat the same steps of A)

For the subinterval [1,1.25]:

f(x)=3^x-x^4\\\\f(1)=3^1-(1)^4=3-1=2\\\\f(1.25)=3^1^.^2^5-(1.25)^4=3.94-2.44=1.5

f(1) and f(1.25) have the same signs, this means there's no zero in the subinterval [1,1.25].

For the subinterval [1.25,1.5]:

f(x)=3^x-x^4\\\\f(1.25)=3^1^.^2^5-(1.25)^4=3.94-2.44=1.5\\\\f(1.5)=3^1^.^5-(1.5)^4=5.19-5.06=0.13

f(1.25) and f(1.5) have the same signs, this means there's no zero in the subinterval [1.25,1.5].

For the subinterval [1.5,1.75]:

f(x)=3^x-x^4\\\\f(1.5)=3^1^.^5-(1.5)^4=5.19-5.06=0.13\\\\f(1.75)=3^1^.^7^5-(1.75)^4=6.83-9.37=(-2.54)

f(1.5) and f(1.75) have opposite signs, this means there's a zero between the subinterval [1.5,1.75].

For the subinterval [1.75,2]:

f(x)=3^x-x^4\\\\f(1.75)=3^1^.^7^5-(1.75)^4=6.83-9.37=(-2.54)\\\\f(2)=3^2-(2)^4=9-16=(-7)

f(1.75) and f(2) have the same signs, this means there isn't a zero between the subinterval [1.75,2].

The graph of the function shows that the answers are correct.

6 0
3 years ago
Let v represent the value of a bank acount, in thousands of dollars, t yearsaftertye account is opend at the bank
Liula [17]

Answer:

could you explain it more

Step-by-step explanation:

4 0
3 years ago
If f(x) = (1/8)(8^x), what is f(3)?
faust18 [17]

Answer:

f(3)=64

Step-by-step explanation:

f(3)=(1/8)(8 cubed)

f(3)= 8 to the negative one times 8 cubed

f(3)= 8 squared

f(3)=64

5 0
3 years ago
Read 2 more answers
What is the solution for the determinant?
Liono4ka [1.6K]
<span>The value of the determinant of a 2x2 matrix is the product of the top-left and bottom-right terms minus the product of the top-right and bottom-left terms.

The value of the determinant of a 2x2 matrix is the product of the top-left and bottom-right terms minus the product of the top-right and bottom-left terms.

= [ (1)(-3)] - [ (7)(0) ]
= -3 - 0
= -3

Therefore, the determinant is -3.

Hope this helps!</span>
4 0
3 years ago
Which expression is equivalent to (16 x Superscript 8 Baseline y Superscript negative 12 Baseline) Superscript one-half?.
loris [4]

To solve the problem we must know the Basic Rules of Exponentiation.

<h2>Basic Rules of Exponentiation</h2>
  • x^ax^b = x^{(a+b)}
  • \dfrac{x^a}{x^b} = x^{(a-b)}
  • (a^a)^b =x^{(a\times b)}
  • (xy)^a = x^ay^a
  • x^{\frac{3}{4}} = \sqrt[4]{x^3}= (\sqrt[3]{x})^4

The solution of the expression is \dfrac{4x^4}{y^6}.

<h2>Explanation</h2>

Given to us

  • (16x^8y^{12})^{\frac{1}{2}}

Solution

We know that 16 can be reduced to 2^4,

=(2^4x^8y^{12})^{\frac{1}{2}}

Using identity (xy)^a = x^ay^a,

=(2^4)^{\frac{1}{2}}(x^8)^{\frac{1}{2}}(y^{12})^{\frac{1}{2}}

Using identity (a^a)^b =x^{(a\times b)},

=(2^{4\times \frac{1}{2}})\ (x^{8\times\frac{1}{2}})\ (y^{12\times{\frac{1}{2}}})

Solving further

=2^2x^4y^{-6}

Using identity \dfrac{x^a}{x^b} = x^{(a-b)},

=\dfrac{2^2x^4}{y^6}

=\dfrac{4x^4}{y^6}

Hence, the solution of the expression is \dfrac{4x^4}{y^6}.

Learn more about Exponentiation:

brainly.com/question/2193820

8 0
2 years ago
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